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Old Oct 14th 2009, 10:38 AM   #1
JAW
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Join Date: Oct 2009
Posts: 4
Pulley on a slope.

I have the following question to answer. Can anyone please help.

A box of mass 150kg rests on a slope that is inclined to the horizontal by 35 degrees. The box is restrained so that it cannot slide down the incline but is free to move up the incline. At the top of the incline a pulley is mounted, over which a rope passes, which is attached to the box at one end and to a water tank at the other end. The water tank is free to move vertically such that if the tank moves down wards, the box moves up the incline. The pulley is positioned so that when the box is at rest the rope is inclined at an angle of 5 degrees above the slope (i.e. 40 degrees to the horizontal. You can assume that the pulley is frictionless, that the rope is light and inextensible and that the coefficent of friction between the surface of the incline and the box is 0.2.

(1) Determine the mass of water that must be added to the tank (assuming that the mass of the tank is 25kg) to just move the box up the incline.

(2) Assuming that the box is repositioned on the incline (same rope angle) and that 5 times the mass of water calculated in part 1 is instantaneously added to the tank, determine the acceleration of the box.


This is what i have so far...

Total R horizontal = T cos 5 - F - mgcos 35 = 0 (1)

Total R Perpendicular = Rn + T sin 5 - mgcos 35 = 0 (2)

F=u Rn (3)

For water tank

R vertical = T -Mwg = 0 (4)

Using 1&3 i can find T

T cos 5 - u Rn - mgsin 35 (5)

from 2 and 4

Rn + (mg sin 35 + u Rn / cos 5) sin 5 - mgcos 35 = 0

Is this the correct way in which to tackle this problem and if so any hints as to where to go with it now...

Thank you
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Old Oct 14th 2009, 09:36 PM   #2
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For (1), the following eqns hold

T = (25 + Mw) g

T cos 5= 150 sin 35 + 0.2 (150 cos 35 - T sin 5).

Find Mw

For (2),

Let T1 be the new tension and a be the accln

(25 + 5 Mw) g - T1 = (25 + 5 Mw) a

T1 cos 5 - [150 sin 35 + 0.2 (150 cos 35 - T sin 5)] = 150 a

Find a
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